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Chapter 2

Q. 2.5.3

An Example of Subsystems: Two Coupled Tanks

Consider two brine tanks each containing 500 L (liters) of brine connected as shown in Figure 2.5.1. At any time t, the first and the second tank contain x_1(t) and x_2(t) kg of salt, respectively. The brine concentration in each tank is kept uniform by continuous stirring. Brine containing r kg of salt per liter is entering the first tank at a rate of 15 L/min, and fresh water is entering the second tank at a rate of 5 L/min. The incoming brine density r(t) can be changed to regulate the process, so r(t) is an input variable.

Brine is pumped from the first tank to the second one at a rate of 60 L/min and from the second tank to the first one at a rate of 45 L/min. Brine is discharged from the second tank at a rate of 20 L/min.

a. Obtain the differential equations, in terms of x_1 and x_2, that describe the salt content in each tank as a function of time.
b. Obtain the transfer functions X_1(s)/R(s) and X_2(s)/R(s).
c. Suppose that r(t) = 0.2 kg/L. Determine the steady-state values of x_1 and x_2, and estimate how long it will take to reach steady state.

Annotation 2022-10-08 231850

Step-by-Step

Verified Solution

a. We assume that the liquid volume does not change when the salt is dissolved in it. We also observe that the total volume of the brine in each tank remains constant at 500 L because the incoming and the outgoing volume flow rates for each tank are equal. Therefore, each liter of brine in the first tank contains x_1/500 kg of salt, and salt leaves the first tank at a rate of 60(x_1/500) kg/min.

Considering that each liter of brine in the second tank contains x_2/500 kg of salt, the rate at which salt leaves the second tank and enters the first one is 45(x_2/500) kg/min. In addition, new brine enters the first tank at a rate of 15r kg/min because each liter of the new brine contains r kg of salt. A similar argument can be given for the second tank. Then the rates of change of the salt content of each tank, in kg/min, can be expressed as

\frac{dx_1}{dt}=15r(t)-60\frac{x_1}{500}+45\frac{x_2}{500} \quad (1) \\ \frac{dx_2}{dt}=0+60\frac{x_1}{500}-65\frac{x_2}{500} \quad (2)

b. Multiplying each equation by 100 gives

100\frac{dx_1}{dt}=1500r(t)-12x_2+9x_2 \\ 100\frac{dx_2}{dt}=12x_1-13x_2

Transforming each equation, using zero initial conditions, and collecting terms gives

(100s+12)X_1(s)-9X_2(s)=1500R(s) \\ -12X_1(s)+(100s+13)X_2(s)=0

Solving these equations gives the transfer functions.

\frac{X_1(s)}{R(s)}=\frac{1500(100s+13)}{10^4s^2+2500s+48} \quad (3) \\ \frac{X_2(s)}{R(s)}=\frac{1.8 \times 10^4}{10^4s^2+2500s+48} \quad (4)

Note that the numerators are different, whereas the denominators are the same.

c. The characteristic roots are the roots of the denominator, which are s = −0.021 and s = −0.029. Thus, the system is stable and the time constants are 𝜏 = 1/0.021 = 47.62 and 𝜏 = 1/0.229 = 4.37. The dominant-time constant is 47.62 minutes, so the time to reach steady state is approximately 4(47.62) = 190.5 minutes. The salt content x_1 may actually take longer than this estimate because of the sterm in the numerator of its transfer function.

Determination of the actual value requires solving (1) and (2) or (3) and (4).

Applying the final-value theorem to (3) and (4) with R(s) = 0.2/s gives the steady-state values.

x_{1ss}=\underset{s\rightarrow 0}{\text{lim}}s \{ [\frac{1500(100s+13)}{10^4s^2+2500s+48}]\frac{0.2}{s} \} =\frac{1500(13)(0.2)}{48}=81.25 \ kg

x_{2ss}=\underset{s\rightarrow 0}{\text{lim}}s \{ [\frac{1.8 \times 10^4}{10^4s^2+2500s+48}]\frac{0.2}{s} \} =\frac{1.8 \times 10^4}{48} 0.2=75 \ kg

These are the steady-state values regardless of the initial values of x_1 and x_2.

This example illustrates a typical problem in system dynamics. The system consists of two subsystems, the two tanks. To model the entire system, we must first understand the dynamics of each subsystem. This was illustrated in Example 2.1.2, where we derive the model of a single tank. Only then can we build the model of the system having two tanks. Building a model of a more complex system, say one containing many tanks, would then proceed in a similar way.